1
Systems of Linear
Equations
Section 1.1 Solutions and Elementary Operations
Practical problems in many fields of study – such as biology, business,
chemistry, computer science, economics, electronics, engineering and
social sciences-can often be reduced to solving a system of linear
equations. Linear Algebra arose from attempts to find systematic method
for solving these systems. Here is the example:
Example 1. A biologist wants to create a diet from fish and meal that contains 183
grams of protein and 93 grams of carbohydrate per day. If fish contains
70% protein and 10% carbohydrate, and meal 30% protein and 60%
carbohydrate, how much of each food is required each day?
Solution
If x grams of fish and y grams of meal are given each day, the amount of
protein and carbohydrate in the diet are represented by these equations:
0.7 x 0.3 y 183
0.1x 0.6 y 93
Multiply the second equation by 7 gives
0.7 x 0.3 y 183
0.7 x 4.2 y 651
Subtracting the top equation from the bottom equation gives 3.9 y
468 ;
then dividing by 3.9 gives y 120 . Substituting this into the equation
0.1x 0.6 y 93 gives x 210 . Hence, the diet consists of 210 grams of
fish and 120 grams of meal per day.
If a, b, c are real numbers, the graph of an equation of the form
ax by
c
is a straight line ( if a, b are not both zero ), so such an equation is called
a linear equation in the variable x and y . However, it is often convenient
to write the variables as x1, x2 , x3 ,..., xn , particularly when more than two
variables are involved. An equation of the form:
a1x1
a2 x2
a3 x3
... an xn
b
is called a linear equation in the n variables x1, x2 , x3 ,..., xn . Here
a1, a2 , a3 ,..., an
denote real numbers ( called the coefficients of
x1, x2 , x3 ,..., xn , respectively ) and b is also a number ( called the constant
term of the equation ). A finite collection of linear equations in the
variables x1, x2 , x3 ,..., xn is called a system of linear equations in these
variables. Hence,
2 x1 3x2
5x3
7
is a linear equation; the coefficient of x1, x2 , x3 are 2, -3,5, respectively,
and the constant term is 7. Note that each variable in a linear equation
occurs to the first power only.
Given
a
linear
equation
a1x1
a2 x2
a3 x3
... an xn
b,
a
sequence s1, s2 , s3 ,..., sn of n numbers is called a solution to the equation
if:
a1s1
that
x1
is,
s1, x2
if
the
s2 ,..., xn
a2 s2
equation
a3s3 ... an sn
is
satisfied
b
when
the
substitutions
sn are made. A sequence of numbers is called a
solution to a system of equations if it is a solution to every equation the
system.
Example 2
Show that, for arbitrary values of s, t ,
x t
s 1
y t
s 2
z
s
w t
is solution to the system :
x 2y
2x y
3z
3z
w
3
w 0
Solution
Simply substitute these values of x, y, z, w in each equation.
x 2y
2x y
3z
3z
w (t s 1) 2(t s 2) 3s t
w 2(t
1) (t s 2) 3 t 0
3
Because both equations are satisfied , it is a solution for all s, t .
The quantities s, t in Example 2 are called parameters, and the
set of solutions, describe in this way, is said to be given in parametric
form and is called the general solution to the system. It turns out that the
solutions to every system of equations ( if there are solutions ) can be
given in parametric form. The following example shows how this happens
in the simplest systems where only one equation in present.
Example 3
Describe all solution to 3x
y
2 z 6 in parametric form.
Solution
Solving the equation for y in terms of x and z , we get y 3x 2 z 6 . If
s and t are arbitrary then, setting x
s , z t , we get solution:
x s
y 3 2t
z t
6
Of course we could have solved for x :
x
Then, if we take y
x
1(p
3
2q 6)
1 (y
3
2z
6)
p, z q , the solutions are represented as follows:
y
p, z q ,
p, q arbitrary
The same family of solutions can “look” quite different!
When only two variables are involved, the solutions to systems of
linear equations can be described geometrically because the graph of a
linear equation ax by
c is a straight line if a and b are not both zero.
More over a point P(s, t ) with coordinates s and t lies on the line if and
only if as bt c , that is when x
s and y t is a solution to the
equation. Hence, the solution to a system of linear equations correspond
to the points P(s, t ) that lie on all the lines in equation.
In particular, if the system consists of just one equation, there
must be infinitely many solutions because there are infinitely many points
on a line. If the system has two equations, there are three possibilities for
the corresponding straight lines:
1. The line intersect in a single point. Then the system has a unique
solution corresponding to that point.
2. The lines are parallel ( and distinct ) and so do not intersect. Then the
system has no solution.
3.
The lines are identical. Then the system has infinitely many solutions,
i.e. one for each point on the ( common ) line.
When three variables are present, the graph of an equation
ax by cz
d can be shown to a plane and so again provides a “picture”
of the set of solution. However, this graphical method has its limitations:
when more than three variables are involved, no physical image of the
graphs ( called hyperplanes ) is possible. It is necessary to turn to a more
“ algebraic” method of solution.
Before describing the method, we introduce a concept that
simplifies the computation involved. Consider the following system:
3x 2 y z w
1
2 x z 2w 0
3 x y 2 z 5w 2
of three equations in four variables. The array of numbers:
3 2
2 0
3 1
1 1 1
1 2 0
2 5 2
occurring in the system is called the augmented matrix of system. Each
row of the matrix consist of the coefficients of the variables ( in order )
from the corresponding equation, together with the constant term. For
clarity, the constants are separated by a vertical line. The augmented
matrix is just a different way of describing the system of equations. The
array of coefficients of the variables
3 2
2 0
3 1
1 1
1 2
2 5
is called the coefficient matrix of the system and
1
0
2
is called the
constant matrix of the system.
Elementary Operations
The algebraic method for solving systems of linear equations is
described as follows. Two such systems are said to be equivalent if each
has the same set of solutions. A system is solved by writing a series of
system, one after the other, each equivalent to the pervious system. Each
of these systems has the same set of solutions as the original one; the
aim is to end up with a system that is easy to solve. Each system in the
series is obtained from the preceding system by a simple manipulation
chosen so that it does not change the set of solutions.
As an illustration, we solve the system x 2y
2 , 2 x 7 y 7 in
this manner. At each stage, the corresponding augmented matrix is
displayed. The original system is:
x 2y
2x y
2
7
1 2 2
2 1 7
First, subtract twice the first equation from the second. The resulting
system is:
x
2y
2
3 y 11
1
0
2
2
3 11
which is equivalent to the original. At this stage we obtain y
multiplying
the second equation by
1
11
3
by
. The result is the equivalent
3
system:
x
2y
y
2
11
3
1 2
0 1
2
11
3
Finally, we subtract twice the second equation from the first to get another
equivalent system.
x
y
16
3
11
3
1 0
0 1
16
3
11
3
Now this system is easy to solve! And because it is equivalent to the
original system, it provides the solution to that system.
Observe that, at each stage, a certain operation is performed on
the system ( and thus on the augmented matrix ) to produce an equivalent
system. The following operations, called elementary operations, can
routinely be performed on systems of linear equations to produce
equivalent systems.
I. Interchange two equations
II. Multiply one equation by a nonzero number
III. Add a multiple of one equation to a different equation.
Theorem 1
Suppose that an elementary operation is performed on a system of linear
equations. Then the resulting system has the same set of solutions as the
original, so the two systems are equivalent.
Elementary operation performed on a system equations produce
corresponding manipulations of the rows of the augmented matrix. Thus,
multiplying a row of a matrix by a number k means multiplying every
entry of the row by k . Adding one row to another row means adding each
entry of that row to the corresponding entry of the other row. Subtracting
two rows is done similarly.
In hand calculations ( and in computer programs ) we usually manipulate
the rows of the augmented matrix rather than the equations. For this
reason we restate these elementary equations for matrices.
The following are called elementary row operations on a matrix :
I. Interchange two rows
II. Multiply one row by a nonzero number
III. Add a multiple of one row to a different row.
In the illustration preceding Theorem 1 these operations led to a matrix of
the form:
1 0 *
0 1 *
where the asterisks represent numbers. In the case of three equations in
the three variables, the goal is to to produce a matrix of the form:
1 0 0 *
0 1 0 *
0 0 1 *
This does not always happen, as we will see in the next section. Here is
an example in which it does happen.
Example 4
Find all solutions to the following system of equations
3x 4 y
2x 3y
4x 3y
z 1
0
z
2
Solution
The augmented matrix of the original system is:
3 4
2 3
4 3
1 1
0 0
1 2
To create a 1 in the upper left corner we could multiply row 1 through by
1
3
. However, the 1 can be obtained without fractions by subtracting row 2
from row 1. The result is:
1 1
2 3
4 3
1 1
0 0
1 2
The upper left 1 is now used to „clean up‟ the first column, that is create
zeros in the other positions in the column. First subtract 2 times row 1
from row 2 to obtain:
1 1
0 1
4 3
1
2
1
1
2
2
Next subtract 4 times row 1 from row 3. The result is:
1
0
0
1
1
1
1
2
5
1
2
6
This completes the work on column 1. We now use the 1 in the second
position of the second row to clean up the second column by subtracting
row 2 from row 1 and adding row 2 to row 3. For convenience, both row
operations are done in one step. The result is
1 0
0 1
0 0
3
2
7
3
2
8
Note that these manipulations did not affect the first column( the second
row has a zero there ), so our previous effort there has not been
undermined. Finally we clean up the third column. Begin by multiplying
row 3 by
1
7
to obtain:
1 0
0 1
0 0
3
2
1
3
2
8
7
Now subtract 3 times row 3 from row 1, and add 2 times row 3 to row 2 to
get:
3
7
2
7
8
7
1 0 0
0 1 0
0 0 1
3,
7
The corresponding equations are x
y
2, z
7
8
7
, which give the
solutions
Exercises 1.1
1. Verify that the following is solution for all values of s, t
x 2 s 12t 13
y s
z
s 3t 3
w t
is a solution of
2 x 5 y 9 z 3w
x 2 y 4z 1
1
2. Find all solutions to the following in all possible parametric form :
a. 3x 2 y 9
c.
b.
d. 4
2a 2b 5
x 2 y 3z 7
2
3
3. Regarding 4 x 2 y 3 as the equation 4 x 2 y
9
0 z 3 in three variables, find all
solutions in parametric form!
4. Write the augmented matrix for each of the following systems of linear equations:
a b 6
b. b c 7
a c 8
x 2y z 4
a. 2 x y z 7
3 y 3z 9
5. Write the system of linear equations that has each of the following augmented
matrix:
1
2
a.
1
2
0
1
1
2
2
1
3
1
2 10
1 6
1 3
1 0
b.
1
4
3
0 2 9
5 3 4
2 3 5
6. A small furniture factory manufactures tables and chairs. It takes 2 hours to
assemble a table and 30 minutes to assemble a chair. Assembly is carried out by 4
workers on the basis of a single 8-hours shift per day. Customers always buy at four
chairs with each table. Find the model of this case as a system of linear equations.
7.
A farmer owns 200 pigs that consume 90 lb of special feed daily. The feed is
prepared as a mixture of corn and soybean meal with the following compositions:
Feedstuff
Pounds per pound of Feedstuff
Calcium
Protein
Fiber
Corn
0.001
0.09
0.02
Soybean meal
0.002
0.60
0.06
The dietary requirements of the pigs are: 1% calcium, 30% protein, 5 % fiber. Find
the model of this case as a system of linear equations !
8. Two product are manufactured by passing sequentially through three machines.
Times per machine
allocated to the two products is 10 hours per day. The
production time per unit of each product are:
Product
Minutes per unit
Machine 1
Machine 2
Machine 3
1
10
6
8
2
5
20
15
Find the model of this case as a system of linear equations !
9. A company produces two products, A and B. The sales volume for product A is 60%
of the total sales of the two products. The two product use the same material, of
which the daily availability is 100 lb. Products A and B use this raw material at the
rates of 2 lb/unit and 4 lb/unit, respectively. Find the model of this case as a system
of linear equations !
10.